How do you differentiate $y=(5/2)^cosx$ ?

Question

1634 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-19T08:57:25

$y' = - (frac(5)(2))^(cos(x)) ln(frac(5)(2)) sin(x)$

We have: $y = (frac(5)(2))^(cos(x))$

This function can be differentiated using the "chain rule".

Let $u = cos(x)$ and $v = (frac(5)(2))^(u)$ :

$Rightarrow y' = u' cdot v'$

$Rightarrow y' = - sin(x) cdot (frac(5)(2))^(cos(x)) cdot ln(frac(5)(2))$

$Rightarrow y' = - (frac(5)(2))^(cos(x)) ln(frac(5)(2)) sin(x)$

How do you differentiate y=(5/2)^cosxy=(5/2)^cosx?